andrew’s leap 2011

Andrew’s Leap 2011

Pancakes With A Problem Steven

Rudich

The chef at our place is sloppy, and when he prepares a stack of

pancakes they come out all different sizes. Therefore, when I

deliver them to a customer, on the way to the table I rearrange them (so that the smallest winds up on

top, and so on, down to the largest at the bottom) by grabbing several

from the top and flipping them over, repeating this (varying the number I flip) as many times as

necessary.

Developing A Notation:Turning pancakes into numbers


1

2

3

4

5


2

3

4

5

1


52341

So how many flips to sort this particular stack?

52341

4 Flips Are Sufficient

12345

52341

43215

23415

14325

Algebraic Representation

52341

X = The smallest number of flips required to sort:

? X ?Upper Bound

Lower Bound


52341


? X 4Upper Bound

Lower Bound


52341


1 X 4Upper Bound

Lower Bound


52341


2 X 4Upper Bound

Lower Bound


52341


3 X 4Upper Bound

Lower Bound

4 Flips Are Necessary

52341

41325

14325

Flip 1 has to put 5 on bottom.Flip 2 must bring 4 to top.


52341


4 X 4Upper Bound

Lower Bound

Upper Bound

Lower Bound

4 X 4

X = 4

5th Pancake Number

P5 = The number of flips required to sort the

worst case stack of 5 pancakes.

? P5 ?Upper Bound

Lower Bound

5th Pancake Number

P5 = The number of flips required to sort the

worst case stack of 5 pancakes.

4 P5 ?Upper Bound

Lower Bound

The 5th Pancake Number: The MAX of the X’s

120

1199

32 . . . . . . .

X1 X2 X3 X119 X120

52341

4

nth Pancake Number

Pn = The number of flips required to sort the

worst case stack of n pancakes.

? Pn ?Upper Bound

Lower Bound

? Pn ?

What bounds can you prove on

Pn?

Bring To Top Method Bring biggest to top. Place it on bottom. Bring next largest to

top. Place second from

bottom. And so on…

Upper Bound On Pn:Bring To Top Method For n

PancakesIf n=1, no work - we are done.Otherwise, flip pancake n to top and then flip it to position n.

Now use: Bring To Top Method For n-1 Pancakes

Total Cost: at most 2(n-1) = 2n –2 flips.

Better Upper Bound On Pn:Bring To Top Method For n

PancakesIf n=2, at most one flip, we are done.Otherwise, flip pancake n to top and then flip it to position n.

Now use: Bring To Top Method

For n-1 Pancakes

Total Cost: at most 2(n-2) + 1 = 2n –3 flips.

Bring to top not always optimal for a particular stack

5 flips, but can be done in 4 flips

32145

52341

23145

41325

14325

? Pn 2n - 3

What bounds can you prove on

Pn?

916

Breaking Apart ArgumentSuppose a stack S contains a pair of adjacent pancakes that will not be adjacent in the sorted stack. Any sequence of flips that sorts stack S must involve one flip that inserts the spatula between that pair and breaks them apart.

916

Breaking Apart ArgumentSuppose a stack S contains a pair of adjacent pancakes that will not be adjacent in the sorted stack. Any sequence of flips that sorts stack S must involve one flip that inserts the spatula between that pair and breaks them apart. Furthermore, this same principle is true of the “pair” formed by the bottom pancake of S and the plate.

n PnSuppose n is even. S contains n pairs that will need to be broken apart during any sequence that sorts stack S.

2468..n1357..

n-1

S

n PnSuppose n is odd. S contains n pairs that will need to be broken apart during any sequence that sorts stack S.

1357..n2468..

n-1

S

Detail: This construction only works when n>3

n Pn 2n - 3

Bring To Top is

within a factor of two of

optimal

n Pn 2n - 3So starting

from ANY stack we can get to

the sorted stack using no more than Pn

flips.

From ANY stack to Sorted Stack in · Pn

Can you see why we can get from the Sorted

stack to ANY stack in · Pn

flips?Reverse the

sequences we use to sort.

From Sorted Stack to ANY stack in · Pn

Can you see why we can

get from ANY stack S to

ANY stack T in · Pn flips?

From ANY Stack S to ANY stack T in Pn

43512

14352

S T

From ANY Stack S to ANY stack T in Pn

RENAME PANCAKES IN YELLOWSO THAT S IS SORTED. THIS FIXES THE YELLOW ORDERING OF THE STACK T.

43512

14352

41235

12345

S T

The Known Pancake Numbers

123456789

10111213

01345

n Pn

789

1011131415

14 1615 1716 1817 19

P18 Is Unknown

18! Orderings of 18 pancakes.

18! = 6,402,373,705,728,000

Burnt Pancake Numbers

123456789

10

1468

101214151718

n Pn

Is This Really Computer Science?

Posed in Amer. Math. Monthly 82 (1) (1975),

“Harry Dweighter” a.k.a. Jacob Goodman

(17/16)n Pn (5n+5)/3

Bill Gates & Christos

Papadimitriou: Bounds For Sorting By

Prefix Reversal. Discrete

Mathematics, vol 27, pp 47-

57, 1979.

(15/14)n Pn (5n+5)/3

H. Heydari & H. I. Sudborough:

On the Diameter of he

Pancake Network. Journal of

Algorithms, vol 25, pp 67-94,

1997.

Pn 18n/11An (18/11)n Upper

Bound for Sorting by Prefix Reversals

B. Chitturi, W. Fahle, Z. Meng, L. Morales, C. O. Shields, I. H.

Sudborough, and W. Voit. Theoretical

Computer Science, to appear

Pancake Network:Definition For n! Nodes

For each node, assign it the name of one of the n! stacks of n pancakes.

Put a wire between two nodes if they are one flip apart.

Network For n=3

123

321

213

312

231

132

Network For n=4

Pancake Network:Message Routing Delay

What is the maximum distance between two nodes in the network?

Pn

Pancake Network:Reliability

If up to n-2 nodes get hit by lightning the network remains connected, even though each node is connected to only n-1 other nodes.

The Pancake Network is optimally reliable for its number

of edges and nodes.

Mutation Distance One

Can flip any consecutive subsequence

TGCGTCCA

TGCCCTGA

Mutation Distance Three!!!

Transforming cabbage into turnip: polynomial algorithm for sorting signed permutations by

reversals.

Sridhar Hannenhalli Pavel A. Pevzner Journal of the ACM 1999

One “Simple” Problem

A host of problems

and applications

at the frontiers of

science

andrew’s leap 2011

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